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Solve for x and y y+2x=1 and y²-2y-3x=3xy-1-2x²​

User Mcvkr
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1 Answer

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Sure, I'd be happy to help you solve this system of equations!

First, let's solve the first equation for x:

x = 1 - 2y

Now, substitute this expression for x into the second equation:

y² - 2y - 3(1 - 2y) + 1 - 2(1 - 2y)² = 3(1 - 2y)

Expand and simplify:

y² - 2y - 3 + 1 - 4y + 4y² = 3

Combine like terms:

y² - 2y - 3 + 4y² = 7

Simplify:

y² - 2y - 3 = 7

Now, we can factor the left-hand side:

(y - 1)(y + 3) = 7

So, we have two possible solutions:

y - 1 = 0 or y + 3 = 0

Solving for y, we get:

y = 1 or y = -3

Now, let's check our solutions. Substituting y = 1 into the first equation, we get:

x = 1 - 2(1) = 0

This is a valid solution, so (x, y) = (0, 1) is a solution to the system.

Substituting y = -3 into the first equation, we get:

x = 1 - 2(-3) = 7

This is also a valid solution, so (x, y) = (7, -3) is another solution to the system.

Therefore, the solutions to the system are (x, y) = (0, 1) and (x, y) = (7, -3).

User Oleg Pasko
by
7.9k points
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